Question

Capacitance of a capacitor made by a thin metal foil is $2 \mu F$. If the foil is folded with paper of thickness $0.15\, mm$, dielectric constant of paper is $2.5$ and width of paper is $400\, mm$, the length of foil will be

• A. 0.34 m
• B. 1.33 m
• C. 13.4 m
• D. 33.9 m

Answer 1

Answer: (D)

Here, let the length be $1$
width is, $b =400\, mm =0.4 \, m$
thickness, $t =0.15 \, mm =15 \times 10^{-5} \, m$
As capacitance for parallel plate capacitor is,
$C = K \epsilon_{0} A / d = K \epsilon_{0}( l \times b ) / t$
$\Rightarrow l = Ct / K \epsilon_{0} b $
$=2 \times 10^{-6} \times 15 \times 10^{-5} /\left(2.5 \times 8.85 \times 10^{-12} \times 0.4\right)$
$=33.91 \,m$